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Chapter 15: Problem 53

Ant Joy Ride. You place your pet ant Klyde (mass \(m )\) on top of a horizontal, stretched rope, where he holds on tightly. The rope has mass \(M\) and length \(L\) and is under tension \(F\) . You start a sinusoidal transverse wave of wavelength \(\lambda\) and amplitude \(A\) propagating along the rope. The motion of the rope is in a vertical plane. Klyde's mass is so small that his presence has no effect on the propagation of the wave. (a) What is Klyde's top speed as he oscillates up and down? (b) Klyde enjoys the ride and begs for more. You decide to double his top speed by changing the tension while keeping the wavelength and amplitude the same. Should the tension be increased or decreased, and by what factor?

Short Answer

Expert verified
Klyde's top speed is \( A\frac{2\pi v}{\lambda} \). Double his speed by increasing tension by a factor of 4.

Step by step solution

01

Understanding the Ant's Motion

Klyde the ant oscillates vertically as a sinusoidal wave travels along the rope. His maximum speed is reached when he passes through the equilibrium position of the wave.
02

Determine Speed from Displacement Function

The vertical displacement of the rope (and Klyde) can be described as a function of position and time \( y(x,t) = A \sin(kx - \omega t) \), where \( k = \frac{2\pi}{\lambda} \) is the wave number and \( \omega = \frac{2\pi v}{\lambda} \) is the angular frequency.
03

Calculate Maximum Speed

Speed is the derivative of displacement with respect to time, so \( v_{max} = \frac{dy}{dt} = A\omega \cos(kx - \omega t) \). The maximum speed occurs when \( \cos(kx - \omega t) = 1 \), giving \( v_{max} = A\omega \).
04

Relate Tension to Wave Velocity

The wave speed \( v \) on the rope is given by \( v = \sqrt{\frac{F}{\mu}} \), where \( \mu = \frac{M}{L} \) is the linear mass density of the rope. The ant's maximum speed depends on this wave speed as \( v_{max} = A\omega = A \frac{2\pi}{\lambda} v \).
05

Change in Tension for Increased Speed

To double Klyde's top speed \( v'_{max} = 2v_{max}\), we need \( v' = 2v \), hence from \( v' = \sqrt{\frac{F'}{\mu}} = 2\sqrt{\frac{F}{\mu}} \). Thus, \( F' = 4F \). The tension should be increased by a factor of 4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sinusoidal Wave
A sinusoidal wave is a type of wave that exhibits a smooth and periodic oscillation, similar to the sine function in mathematics. This wave propagates in space and time, creating a repetitive motion that is easily recognizable.
A sinusoidal wave is characterized by two main features:
  • **Wavelength (\( \lambda \))**: The distance between consecutive crests or troughs.
  • **Amplitude (\( A \))**: The maximum displacement from the wave鈥檚 equilibrium position.
In our context with Klyde, the introduction of a sinusoidal wave along the rope causes Klyde to experience up-and-down movements. As the wave travels horizontally, Klyde oscillates vertically due to the transverse nature of the wave, where the motion is perpendicular to the direction of the wave's propagation.
Wave Speed
Wave speed is the rate at which the wave propagates along the rope. It plays a crucial role in determining Klyde's maximum oscillation speed. The speed of a wave on a string, like our rope with Klyde, depends on two factors:
  • The **tension (\( F \))** in the rope.
  • The **linear mass density (\( \mu \))** of the rope, defined as the mass per unit length \( (\mu = \frac{M}{L}) \).
The wave speed \( v \) can be calculated using the formula \( v = \sqrt{\frac{F}{\mu}} \). Increasing the tension in the rope increases the wave speed, which directly affects Klyde鈥檚 vertical speed as he passes through the equilibrium position. By changing the wave speed, we can manipulate how fast Klyde moves up and down.
Oscillation
Oscillation refers to the repetitive back-and-forth movement experienced by Klyde as the wave passes beneath him. In this scenario, Klyde's motion is harmonic and occurs in a regular, cyclic pattern.
The speed of oscillation relates closely to the wave's properties. It is calculated by deriving the wave's displacement function \( y(x,t) = A \sin(kx - \omega t) \) with respect to time. This derivative gives Klyde's vertical speed \( v_{max} = A\omega \cos(kx - \omega t) \), reaching its maximum when \( \cos(kx - \omega t) = 1 \), i.e., \( v_{max} = A\omega \). Here, \( \omega \) is the angular frequency \( (\omega = \frac{2\pi v}{\lambda}) \), which links oscillation speed with wave speed. Adjustments to tension affect angular frequency \( \omega \), thereby altering oscillation characteristics.
Tension in Rope
Tension in the rope significantly impacts both the wave speed and the behavior of sinusoidal waves traveling along the rope. This tension creates a restoring force that helps the wave propagate.
In our scenario, the tension is vital for determining Klyde鈥檚 oscillation speed. The relationship between tension and wave speed is expressed through the equation \( v = \sqrt{\frac{F}{\mu}} \). Increasing tension resultantly increases the wave speed, thereby doubling Klyde's oscillation speed when the tension is increased by a factor of four, as shown in the exercise solution. This is due to the direct relationship between tension and velocity. Changing tension allows for control over the dynamic properties of waves on the string, affecting Klyde's ride.

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Most popular questions from this chapter

(a) Show that for a wave on a string, the kinetic energy per unit length of string is $$ u_{k}(x, t)=\frac{1}{2} \mu v_{y}^{2}(x, t)=\frac{1}{2} \mu\left(\frac{\partial y(x, t)}{\partial t}\right)^{2} $$ where \(\mu\) is the mass per unit length. (b) Calculate \(u_{\mathrm{k}}(x, t)\) for a sinusoidal wave given by Eq. \((15.7) .\) (c) There is also elastic potential energy in the string, associated with the work required to deform and stretch the string. Consider a short segment of string at position \(x\) that has unstretched length \(\Delta x,\) as in \(\mathrm{Fg} .15 .13 .\) Ignoring the (small) curvature of the segment, its slope is \(\partial y(x, t) / \partial x\) Assume that the displacement of the string from equilibrium is small, so that \(\partial y / \partial x\) has a magnitude much less than unity. Show that the stretched length of the segment is approximately $$ \Delta x\left[1+\frac{1}{2}\left(\frac{\partial y(x, t)}{\partial x}\right)^{2}\right] $$ (Hint: Use the relationship \(\sqrt{1+u} \approx 1+\frac{1}{2} u,\) valid for \(|u| \ll 1 . )\) (d) The potential energy stored in the segment equalsthe work done by the string tension \(F\) (which acts along the string) to stretch the segment from its unstretched length \(\Delta x\) to the length calculated in part (c). Calculate this work and show that the potential energy per unit length of string is $$ u_{p}(x, t)=\frac{1}{2} F\left(\frac{\partial y(x, t)}{\partial x}\right)^{2} $$ (e) Calculate \(u_{\mathrm{o}}(x, t)\) for a sinusoidal wave given by Eq. \((15.7)\) (f) Show that \(u_{k}(x, t)=u_{p}(x, t)\) for all \(x\) and \(t .(g)\) Show \(y(x, t)\) \(u_{k}(x, t),\) and \(u_{p}(x, t)\) as functions of \(x\) for \(t=0\) in one graph with all three functions on the same axes. Explain why \(u_{k}\) and \(u_{p}\) are maximum where \(y\) is zero, and vice versa. (h) Show that the instantaneous power in the wave, given by Eq. \((15.22),\) is equal to the total energy per unit length multiplied by the wave speed \(v\) Explain why this result is reasonable.

A wire with mass 40.0 \(\mathrm{g}\) is stretched so that its ends are tied down at points 80.0 \(\mathrm{cm}\) apart. The wire vibrates in its fundamental mode with frequency 60.0 \(\mathrm{Hz}\) and with an amplitude at the antinodes of 0.300 \(\mathrm{cm}\) . (a) What is the speed of propagation of transverse waves in the wire? (b) Compute the tension in the wire. (c) Find the maximum transverse velocity and acceleration of particles in the wire.

One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates the rope transversely at 120 \(\mathrm{Hz}\) . The other end passes over a pulley and supports a \(1.50-\mathrm{kg}\) mass. The linear mass density of the rope is 0.0550 \(\mathrm{kg} / \mathrm{m}\) . (a) What is the speed of a transverse wave on the rope? (b) What is the wavelength? (c) How would your answers to parts (a) and (b) change if the mass were increased to 3.00 \(\mathrm{kg}\) ?

A transverse wave on a rope is given by $$ y(x, t)=(0.750 \mathrm{cm}) \cos \pi\left[\left(0.400 \mathrm{cm}^{-1}\right) x+\left(250 \mathrm{s}^{-1}\right) t\right] $$ (a) Find the amplitude, period, frequency, wavelength, and speed of propagation. (b) Sketch the shape of the rope at these values of t: \(0,0.0005\) s, 0.0010 s. (c) Is the wave traveling in the \(+x-\) or \(-x\) -direction? (d) The mass per unit length of the rope is \(0.0500 \mathrm{kg} / \mathrm{m} .\) Find the tension. (e) Find the average power of this wave.

Transverse waves on a string have wave speed 8.00 \(\mathrm{m} / \mathrm{s}\) , amplitude \(0.0700 \mathrm{m},\) and wavelength 0.320 \(\mathrm{m}\) . The waves travel in the \(-x\) -direction, and at \(t=0\) the \(x=0\) end of the string has its maximum upward displacement. (a) Find the frequency, period, and wave number of these waves, (b) Write a wave function describing the wave. (c) Find the transverse displacement of a particle at \(x=0.360 \mathrm{m}\) at time \(t=0.150 \mathrm{s}\) s. (d) How much time must elapse from the instant in part (c) until the particle at \(x=0.360 \mathrm{m}\) next has maximum upward displacement?
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